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Chapter 6: Problem 26

A survey of 15,000 American adults in 2011 found that \(35.3 \%\) identify as Democrats and \(34.0 \%\) identify as Republicans, with the rest identifying as independent or other. \(^{8}\) If we want \(95 \%\) confidence, what is the margin of error in the estimate for the proportion of Democrats? For the proportion of Republicans? Do you feel comfortable concluding that in 2011 more American adults self-identified as Democrats than self-identified as Republicans? Explain.

Short Answer

Expert verified
The margin of error for Democrats is approximately 0.54%. And for Republicans, it's approximately 0.53%. Even though the proportion of Democrats in the sample is higher than that of Republicans, we can't confidently state that there were more Democrats than Republicans amongst American adults in 2011, since the confidence intervals overlap.

Step by step solution

01

Define the problem clearly

To estimate a population proportion with a certain level of confidence, we need to calculate the margin of error. The formula for computing the margin of error is \(E=Z_{\alpha/2}*鈭歿[ P(1-P) / n ]}\) where E is margin of error, P is the proportion, n is the total size of the sample, and Z is the critical value for the required level of confidence (in this case, Z value for 95% confidence is 1.96).
02

Calculate the margin of error for Democrats

First, convert the proportion to a decimal form. For Democrats, the proportion P is 0.353. The sample size (n) given in the problem is 15000. Z is 1.96 for 95% confidence level. Plugging these values into the margin of error formula, we get \(E=1.96*鈭歿[ 0.353(1-0.353) / 15000 ]}\). After calculating, the margin of error for the Democrats is approximately 0.0054 or 0.54%.
03

Calculate the margin of error for Republicans

We repeat the calculation with the proportion for Republicans, which is 0.34. Inserting the values into the margin of error formula, we get \(E=1.96*鈭歿[ 0.34(1-0.34) / 15000 ]}\). After performing the calculation, the margin of error for Republicans is approximately 0.0053 or 0.53%.
04

Comparing the confidence intervals

The confidence interval for Democrats is \(0.353 \pm 0.0054\) and for Republicans it is \(0.34 \pm 0.0053\). These intervals overlap, meaning that it is possible that the true population proportion is within both intervals.
05

Conclusion

Even though more people in the sample identify as Democrats than Republicans, due to the overlapping confidence intervals we cannot confidently conclude that more American adults self-identified as Democrats than self-identified as Republicans in 2011. The difference in the sample could be due to sampling variability.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Population Proportion
When conducting a survey, researchers aim to learn about a larger group, known as a population, by examining a subset of that group, or a sample. The population proportion is the fraction of the population that has a certain characteristic. For instance, in a political survey, the population proportion would refer to the percentage of all individuals in the population that identify with a specific political party.

In the given exercise, the goal is to understand what portion of the American adult population identifies as Democrats or Republicans. It is crucial to note that we can never be entirely certain of the exact population proportion without surveying every single individual, which is impractical. This is why we work with a sample and use it to estimate the population proportion within a certain range of uncertainty, typically expressed as a margin of error in percentage points.
Confidence Interval
A confidence interval provides a range of values, derived from the sample, which is likely to contain the population proportion. It is constructed around a sample proportion and extends on either side by the margin of error. Confidence intervals give us a sense of how precise our estimate of the population proportion is.

The 95% confidence interval used in our exercise means that if we were to draw numerous samples and calculate the confidence interval for each, we would expect that approximately 95% of these intervals will contain the true population proportion. The width of the confidence interval is affected by the margin of error and the level of confidence we desire. A narrower interval indicates more precision but less confidence that the interval contains the true proportion, while a wider interval implies less precision but increased confidence.
Sampling Variability
Sampling variability, also referred to as sampling error, represents the variation that naturally occurs from one sample to another, simply due to the randomness inherent in the sampling process. It is an important concept because it influences the margin of error and, consequently, the width of our confidence interval.

In our political survey example, the variability might occur if one sample had slightly more Democrats by chance, while another had slightly more Republicans. This natural fluctuation means that different samples will likely yield different estimates of the population proportion. Recognizing sampling variability helps us understand that the observed differences between sample proportions may not reflect actual differences in the population but could be due to the random nature of sampling.

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Most popular questions from this chapter

When we want \(95 \%\) confidence and use the conservative estimate of \(p=0.5,\) we can use the simple formula \(n=1 /(M E)^{2}\) to estimate the sample size needed for a given margin of error ME. In Exercises 6.40 to 6.43, use this formula to determine the sample size needed for the given margin of error. A margin of error of 0.02 .

A survey of 1000 adults in the US conducted in March 2011 asked "Do you favor or oppose 'sin taxes' on soda and junk food?" The proportion in favor of taxing these foods was \(32 \% .10\) (a) Find a \(95 \%\) confidence interval for the proportion of US adults favoring taxes on soda and junk food. (b) What is the margin of error? (c) If we want a margin of error of only \(1 \%\) (with \(95 \%\) confidence \()\), what sample size is needed?

A data collection method is described to investigate a difference in means. In each case, determine which data analysis method is more appropriate: paired data difference in means or difference in means with two separate groups. A study investigating the effect of exercise on brain activity recruits sets of identical twins in middle age, in which one twin is randomly assigned to engage in regular exercise and the other doesn't exercise.

Quiz Timing A young statistics professor decided to give a quiz in class every week. He was not sure if the quiz should occur at the beginning of class when the students are fresh or at the end of class when they've gotten warmed up with some statistical thinking. Since he was teaching two sections of the same course that performed equally well on past quizzes, he decided to do an experiment. He randomly chose the first class to take the quiz during the second half of the class period (Late) and the other class took the same quiz at the beginning of their hour (Early). He put all of the grades into a data table and ran an analysis to give the results shown below. Use the information from the computer output to give the details of a test to see whether the mean grade depends on the timing of the quiz. (You should not do any computations. State the hypotheses based on the output, read the p-value off the output, and state the conclusion in context.) Two-Sample T-Test and Cl \begin{tabular}{lrrrr} Sample & \(\mathrm{N}\) & Mean & StDev & SE Mean \\ Late & 32 & 22.56 & 5.13 & 0.91 \\ Early & 30 & 19.73 & 6.61 & 1.2 \\ \multicolumn{3}{c} { Difference } & \(=\mathrm{mu}(\) Late \()\) & \(-\mathrm{mu}\) (Early) \end{tabular} Estimate for difference: 2.83 $$ \begin{aligned} &95 \% \mathrm{Cl} \text { for difference: }(-0.20,5.86)\\\ &\text { T-Test of difference }=0(\text { vs } \operatorname{not}=): \text { T-Value }=1.87\\\ &\text { P-Value }=0.066 \quad \mathrm{DF}=54 \end{aligned} $$

Effect of Splitting the Bill Exercise 2.153 on page 105 describes a study to compare the cost of restaurant meals when people pay individually versus splitting the bill as a group. In the experiment half of the people were told they would each be responsible for individual meal costs and the other half were told the cost would be split equally among the six people at the table. The 24 people paying individually had a mean cost of 37.29 Israeli shekels with a standard deviation of 12.54 , while the 24 people splitting the bill had a higher mean cost of 50.92 Israeli shekels with a standard deviation of 14.33. The raw data can be found in SplitBill and both distributions are reasonably bell-shaped. Use this information to find and interpret a \(95 \%\) confidence interval for the difference in mean meal cost between these two situations.
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